$\left\|A\right\|_{2}\le 1$. This can be shown using some integral inequalities and Holder's inequality. I have not been able to show that this bound is tight, unfortunately.
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Nick ThompsonJun 9 '12 at 5:18

I suspect it might be easier to look at $A$ in terms of the basis $e_n(t) = e^{i n 2\pi t}$.
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copper.hatJun 9 '12 at 7:00

It is Problem 188 in the book by P. Halmos, "A Hilbert space problem book". In the solution, the author writes that "A direct approach seems to lead nowhere." The norm is indeed $2/\pi$, and is computed through the adjoint $A^*$ and a suitable kernel. It is a rather long proof, so please try to read it on Halmos' book.